How to translate CNOT gate by Hadamard and Pauli Z gate in matrix form?

Question

I am trying to understand how CNOT or X gate can be explained by even more basic gates.

I went through these 2 links that explained the relation, but for my understanding, I am trying to prove it by matrix form. And yes I am clearly disabled in mathematics.

Link:1 Link:2

So, let's take the case where first Qbit(Q1) is the control and the second Qbit(Q2) is the target.

As per the articles, CNOT can be explained by $(I\times H)\cdot Z\cdot(I\times H)$. where X is matrix multiplication and (.) refers to dot product.

So, I went on to first convert all of them to the matrix Step 1:

$\left(\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\times{1\over\sqrt2}\begin{bmatrix}1 & 1\\1 & -1\end{bmatrix}\right)\cdot\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}\cdot\left(\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\times{1\over\sqrt2}\begin{bmatrix}1 & 1\\1 & -1\end{bmatrix}\right)$

I have my doubts if this understanding itself is correct but let's continue hoping it is.

Step 2: $\left({1\over\sqrt2}\begin{bmatrix}1 & 1 & 0 & 0\\1 & -1 & 0 & 0\\0 & 0 & 1 & 1\\0 & 0 & 1 & -1 \end{bmatrix}\right)\cdot\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}\cdot\left({1\over\sqrt2}\begin{bmatrix}1 & 1 & 0 & 0\\1 & -1 & 0 & 0\\0 & 0 & 1 & 1\\0 & 0 & 1 & -1 \end{bmatrix}\right)$

Does this approach look right? if so, how should I approach further? if not what am I missing?

I even tried to take a smaller bit to solve and tried to solve this:

$X = H\cdot Z \cdot H$

Step 1: $X = {1\over\sqrt2}\begin{bmatrix}1 & 1\\1 & -1\end{bmatrix}\cdot \begin{bmatrix}1 & 0\\0 & -1\end{bmatrix} \cdot {1\over\sqrt2}\begin{bmatrix}1 & 1\\1 & -1\end{bmatrix}$

Step 2: $X = {1\over2}\begin{bmatrix} 1*1 & 1*0 \\ 1*0 & -1*-1\end{bmatrix}\cdot \begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix}$

Step 3: $X = {1\over2}\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}\cdot \begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix}$

Step 4: $X = {1\over2}\begin{bmatrix} 1*1 & 0*1 \\ 0*1 & 1*-1\end{bmatrix}$

Step 5: $X = {1\over2}\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}$

once again, this is not Pauli-X gate, so I must be doing something stupid. Please help and guide me.

Answer 1

I think you have a slight misunderstanding of what the $Z$ gate in the screenshot you attached is. It is, in fact, not a Z gate, but a (2-qubit) $CZ$ or controlled-$Z$. gate, also referred to as a controlled-phase gate (because the $Z$ operation is a flip of the phase of the $|1\rangle$ state).

This $CZ$ gate is a lot like the $CX$ gate, however, it performs a $Z$ operation on the second qubit, but only if the state of the first qubit is $|1\rangle$ state. If the first is in the $|0\rangle$ state it performs nothing (i.e. $I$ operation) on the second qubit instead.

In more technical terms, we can write thus: \begin{equation} CZ = |0\rangle \langle0| \otimes I + |1\rangle \langle1| \otimes Z. \end{equation}

Then: \begin{equation} (I \otimes H) CZ(I\otimes H) = |0\rangle \langle0| \otimes HIH + |1\rangle \langle1| \otimes HZH. \end{equation}

We can simplify the $HIH$ term easily: $HIH = HH = I$, from the properties of $I$ and $H$. The $HZH$ term is actually equal to $X$: \begin{equation} \begin{split} HZH &= \frac{1}{2}\begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix}\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}\begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix}\\ &= \frac{1}{2}\begin{bmatrix}1 & -1 \\ 1 & 1\end{bmatrix}\begin{bmatrix}1 & 1 \\ 1 & -1\end{bmatrix} \\ &= \frac{1}{2}\begin{bmatrix}0 & 2 \\ 2 & 0\end{bmatrix}\\ &= \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix} = X. \end{split} \end{equation} So we can write: \begin{equation} (I \otimes H) CZ(I\otimes H) = |0\rangle \langle0| \otimes HIH + |1\rangle \langle1| \otimes HZH = |0\rangle \langle0| \otimes I + |1\rangle \langle1| \otimes X. \end{equation} That is to say, an operation that performs the $X$ operation on the second qubit if the first qubit is in the $|1\rangle$ state (and nothing if the first qubit is in the $|0\rangle$ state). This is, of course, exactly the $CX$ gate which we are looking for.